Lesson Video: Elasticity and Kinematics

In this video we learn how elasticity and elastic potential energy are related to projectile motion described using the kinematic equations.


Video Transcript

In this video, we’re going to learn about elasticity and kinematics. We’ll see how these two areas are connected. And we’ll get some practice solving problems that involve them both.

To start out, imagine that, one day, as you’re playing with a ball, you kick it up to a high enough height that it becomes stuck in the branches of a tree. In order to get the ball back, you take out your trusty slingshot and load it with a small pebble. You aim to shoot the pebble so that it hits the ball and dislodges it from the tree. To do this, you’ll need to very precisely aim and also pull back on the slingshot.

To understand better how to do something like this, it’s helpful to know about elasticity and kinematics. Elasticity in general is the ability a material has to return to its original shape after it’s been stretched or compressed. Objects like a rubber band, a spring, or a bungee cord all exhibit this property of elasticity.

Not only does an elastic object work to restore itself to its original shape after being stretched or compressed, but we know that, in the case of springs, that restored force is proportional to the amount we’ve stretched or compressed it multiplied by the spring constant 𝑘.

The fact that an elastic object when its shape is changed exerts a force to try to restore itself to its original shape means that the object is capable of doing work. And that means that there is elastic potential energy stored in such an object proportional to the spring constant of that object times its displacement from equilibrium squared.

We know that potential energy has the capacity to be turned into kinetic energy or energy of motion. For example, imagine that we have a spring-loaded tennis ball launcher. If we load the launcher with a ball and compress the spring a distance Δ𝑥 in the process, then given the spring constant 𝑘, our tennis ball now has potential energy. And if we release the spring inside the launcher, the ball itself is released. And our potential energy turns into kinetic energy.

As soon as the ball leaves the launcher, it becomes a projectile, moving according to the kinematic equations of motion. That is, our object of mass 𝑚 is given a speed 𝑣 because of our elastic potential energy. The kinematic equations describe the motion of projectiles, things flying through the air or even sliding along the ground at a constant acceleration.

So elasticity has the capacity to create elastic potential energy. When this potential energy is released, objects are put in motion. And the kinematic equations of motion can be a helpful tool for describing that. Let’s get some practice combining elasticity and kinematics through a couple of examples.

In the reality television show “Amazing Race,” a contestant is firing 12-kilogram watermelons from a slingshot to hit targets down the field. The slingshot is pulled back 1.5 meters and the watermelon is at ground level. The launch point is 0.3 meters above ground. And the target is 10 meters away horizontally from the launch point. Calculate the spring constant of the slingshot.

We can call the spring constant of the slingshot 𝑘. And we’ll start out by drawing a diagram of the situation. In this scenario, we have our watermelons of mass we’ve called 𝑚, 12 kilograms, being loaded into an elastic slingshot. The slingshot has been pulled back a total distance Δ𝑥 of 1.5 meters. And when the slingshot is released, the watermelon is launched from an initial height we’ve called ℎ of 0.3 meters. If it’s aimed on target, the watermelon then travels a horizontal range we’ve called 𝑅 of 10 meters before it hits the target. Based on this information, we want to solve for the spring constant of this elastic slingshot.

We know that when it’s released, the slingshot will give the watermelon kinetic energy. And because it raises the watermelon a height ℎ, it also gives that watermelon potential energy due to gravity. When we recall that elastic potential energy equals one-half 𝑘, the spring constant, times the displacement, Δ𝑥, squared, that gravitational potential energy equals mass times 𝑔 times ℎ, where we’ll treat 𝑔, the acceleration due to gravity, as exactly 9.8 meters per second squared.

And we recall that an object’s kinetic energy is half its mass times its speed squared. Then we can write that one-half the spring constant, 𝑘, times the displacement, Δ𝑥, squared is equal to 𝑚 times 𝑔 times ℎ, where 𝑚 is the mass of the watermelon and ℎ is its launch height, plus one-half the watermelon’s mass times its launch speed, 𝑣, squared.

If we rearrange this equation to solve for 𝑘, we see that we know the mass of the watermelon 𝑚, we know the displacement of the slingshot Δ𝑥, we know the acceleration due to gravity 𝑔, the height ℎ. But the speed 𝑣 of the watermelon when it’s launched is the one thing we don’t yet know. That means if we can solve for 𝑣, then we can solve for 𝑘. So let’s figure out what that launch speed 𝑣 of the watermelon is.

To do that, let’s start by dividing the motion of our watermelon projectile into vertical, 𝑦, and horizontal, 𝑥, directions. If we call the total time that our watermelon is in flight before it reaches our target 𝑡, then we can write that 𝑣 sub 𝑥, the horizontal speed of our projectile, is equal to the range, 𝑅, over the time it travels, 𝑡.

We’ve called 𝑣 the launch speed of our watermelon at its initial launch point. And if we look at the right triangle formed by the slingshot, the ground, and the support post, we can see that the horizontal speed of our object is equal to its overall initial speed, 𝑣, multiplied by the cosine of the angle that we’ve called 𝜃, the angle between the slingshot and the ground.

Looking at this equation, we can solve for 𝜃 because we know the height ℎ as well as Δ𝑥. And we’re given the range 𝑅. But we don’t know the total time in flight 𝑡, which will let us solve then for 𝑣, the initial launch speed of the watermelon. So let’s now move to considering motion in the 𝑦 or vertical direction to see if that can help us solve for the overall time, 𝑡.

In the vertical direction, our projectile experiences a steady acceleration due to gravity. And therefore, its motion is described by the kinematic equations. Say we pick two points along the watermelon’s trajectory. The first point is at its launch point. And the second is at the zenith of its trajectory, where its altitude is highest. And we’ll call these points 𝑖 and 𝑓 for initial and final.

There is a kinematic equation which says that final speed is equal to initial speed plus acceleration times time passed. If we apply that particular equation to our chosen initial and final points, we can write that 𝑣 sub 𝑓 is equal to 𝑣 sub 𝑖 minus 𝑔 times one-half the total time in flight 𝑡.

Recalling that we’re only considering vertical motion at the moment, we know that the vertical speed of our projectile at point 𝑓 is zero. It’s neither going up nor down. So 𝑣 sub 𝑓 in our equation is zero. And we can write that 𝑣 sub 𝑖 is equal to 𝑔 times 𝑡 over two. 𝑣 sub 𝑖 is the initial speed of our projectile in the 𝑦 or vertical direction.

Based on our diagram, we can write 𝑣 sub 𝑖 as the overall launch speed, 𝑣, multiplied by the sin of the angle 𝜃. When we rearrange this equation to solve for 𝑡, we find it’s equal to two times 𝑣 times the sin of 𝜃 all over 𝑔.

What we’ll do now is take this expression for the overall time in flight 𝑡 and substitute it in for 𝑡 in our equation for 𝑣 times the cos of 𝜃. When we do, we find that 𝑣 cos 𝜃 equals 𝑅𝑔 over two 𝑣 sin 𝜃. And when we multiply both sides by the denominator, we have an expression on the left-hand side of our equation that we can simplify through a trigonometric identity. Two times the cosine of an angle times the sine of that same angle is equal to the sine of twice that angle. So we can write that 𝑣 squared times the sin of two 𝜃 is equal to 𝑅 times 𝑔. Or solving for 𝑣, it’s equal to the square root of 𝑅 times 𝑔 over the sin of two 𝜃.

We now have an expression for 𝑣 that we can plug into our original energy balance equation to help us solve for the spring constant 𝑘. When we do plug in this term, we see we can factor out 𝑔 from this equation. And looking at this simplified equation, we see we have values for 𝑚, 𝑔, Δ𝑥, ℎ, and 𝑅.

The last thing to solve for before calculating 𝑘 is 𝜃. Looking at the right triangle, where we’ve defined 𝜃, we see that the sine of this angle is equal to ℎ over Δ𝑥, both of which are known. If we take the arcsine of both sides of the equation, we find that 𝜃 is equal to the inverse sin of ℎ, 0.3 meters, divided by Δ𝑥, 1.5 meters. Plugging this in on our calculator, we find that 𝜃 is approximately 11.54 degrees.

Now we’re ready to plug in for our variables and solve for 𝑘. When we enter this expression on our calculator, we find that, to two significant figures, 𝑘 equals 1400 newtons per meter. That’s the value of the spring constant of the slingshot.

Now let’s look at another example of elasticity involving energy conservation.

Two 8.00-kilogram masses are connected to each other by a spring with a force constant of 25.0 newtons per meter and a rest length of 1.00 meters. The spring has been compressed to 0.900 meters in length. The masses travel toward each other. And each mass moves at 0.500 meters per second. What is the total energy in the system?

We can label the total system energy we want to solve for capital 𝐸 and begin by drawing a sketch of the situation. In this scenario, we have two identical masses — we’ve labeled them 𝑚 — of value 8.00 kilograms, connected by a spring of spring constant 𝑘. The spring started out at a length we’ve called 𝐿 sub 𝑖 for its initial length 1.00 meters. And now it’s at a length we’ve called 𝐿 sub 𝑓, 0.900 meters. The two masses move towards one another at speeds 𝑣, where 𝑣 is given as 0.500 meters per second. Knowing all this, we want to solve for the total energy in the system at this moment in time.

We know that this total energy will be equal to the sum of the system’s kinetic plus its potential energy. We recall that kinetic energy, in general, is equal to one-half an object’s mass times its speed squared. And since we have two identical masses, 𝑚, moving at identical speeds, 𝑣, we can rewrite this expression for kinetic energy simply as 𝑚 times 𝑣 squared.

When it comes to the potential energy of our system, we have a spring which has been compressed from its natural length. The potential energy due to an elastic deformation is equal to one-half a spring’s constant, 𝑘, multiplied by its displacement from equilibrium, Δ𝑥, squared. In our case, we can write this elastic potential energy as one-half the force constant, 𝑘, times the quantity 𝐿 sub 𝑖 minus 𝐿 sub 𝑓 squared.

In the problem statement, we’re given 𝑚, 𝑣, 𝑘, 𝐿 sub 𝑖, and 𝐿 sub 𝑓. So we’re ready to plug in and solve for 𝐸. When we enter in the values for 𝑚, 𝑣, 𝑘, 𝐿 sub 𝑖, and 𝐿 sub 𝑓 and enter this expression on our calculator, we find that, to three significant figures, 𝐸 is 2.13 joules. That’s the total energy of the system, kinetic plus potential.

Let’s summarize what we’ve learned about elasticity and kinematics. We’ve seen that elasticity can create elastic potential energy, where that elastic potential energy is equal to one-half the force constant or spring constant multiplied by the displacement from equilibrium squared. We’ve also seen that when potential energy is converted to kinetic energy, objects are set in motion. And when those objects move as projectiles, we’ve seen that their motion is described by the kinematic equations of motion.

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